Let a and b be the vectors along the diagonal of a parallelogram having area 2√2. Let the angle betw

Q: Wait, why is the area $\frac{1}{2} |\vec{a} \times \vec{b}|$? I thought the area was just $|\vec{a} \times \vec{b}|$.

That is a very common confusion! If $\vec{a}$ and $\vec{b}$ were the *sides* of the parallelogram, the area would indeed be $|\vec{a} \times \vec{b}|$. But here, they are the *diagonals*. The area of a parallelogram is half the magnitude of the cross product of its diagonals. Think of it as the sum of the areas of the four triangles formed by the intersection of the diagonals.

Q: Why does the term $\vec{b} \cdot (\vec{a} \times \vec{b})$ become zero?

This is the beauty of the cross product! By definition, the vector $\vec{a} \times \vec{b}$ is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$. Since $\vec{b}$ lies in that plane, it must be perpendicular to $\vec{a} \times \vec{b}$. The dot product of any two perpendicular vectors is zero. It is not just math; it is geometric necessity.

Q: I got $\cos \alpha = -\frac{1}{\sqrt{2}}$. Does this mean the angle is negative?

Great catch! The cosine value is negative, which tells us that the angle $\alpha$ is obtuse (greater than $90^{\circ}$). Since $\cos(135^{\circ}) = -\frac{1}{\sqrt{2}}$, the angle is $\frac{3\pi}{4}$. We never reject the negative sign in cosine; it tells us the direction of the vector relative to the reference!

Question

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonal of a parallelogram having area $2\sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. $|\vec{a}| = 1$ and $|\vec{a} \cdot \vec{b}| = |\vec{a} 	imes \vec{b}|$. If $\vec{c} = 2\sqrt{2}(\vec{a} 	imes \vec{b}) - 2\vec{b}$, then an angle between $\vec{b}$ and $\vec{c}$ is:

Options:

(A) $\frac{\pi}{4}$

(B) $\frac{\pi}{4}$

(C) $\frac{5\pi}{6}$

(D) $\frac{3\pi}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space, looking at a parallelogram defined by its diagonals, $\vec{a}$ and $\vec{b}$. The area of a parallelogram is related to its diagonals by the formula:

$$	ext{Area} = \frac{1}{2} |\vec{a} 	imes \vec{b}|$$

Given that the area is $2\sqrt{2}$, we immediately find our first anchor point:

$$|\vec{a} 	imes \vec{b}| = 4\sqrt{2}$$

### The Trigonometric Harmony

We are given the condition $|\vec{a} \cdot \vec{b}| = |\vec{a} 	imes \vec{b}|$. Let $	heta$ be the angle between the diagonals.

Using the definitions of the dot and cross products, we have:

$$|\vec{a}| |\vec{b}| \cos 	heta = |\vec{a}| |\vec{b}| \sin 	heta$$

Since the magnitudes $|\vec{a}|$ and $|\vec{b}|$ are non-zero, they cancel out, leaving $\cos 	heta = \sin 	heta$. Given that the angle is acute, we conclude:

$$	heta = \frac{\pi}{4}$$

### The Construction of $\vec{c}$

The problem introduces a new vector, $\vec{c} = 2\sqrt{2}(\vec{a} 	imes \vec{b}) - 2\vec{b}$. This vector is a linear combination of the cross product (perpendicular to the plane) and the diagonal $\vec{b}$ (within the plane).

To find the angle $\alpha$ between $\vec{b}$ and $\vec{c}$, we use the dot product formula:

$$\cos \alpha = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$$

### The Vanishing Act

Let us calculate the numerator, $\vec{b} \cdot \vec{c}$. Substituting the definition of $\vec{c}$:

$$\vec{b} \cdot [2\sqrt{2}(\vec{a} 	imes \vec{b}) - 2\vec{b}] = 2\sqrt{2}(\vec{b} \cdot (\vec{a} 	imes \vec{b})) - 2|\vec{b}|^2$$

Because $\vec{a} 	imes \vec{b}$ is perpendicular to the plane containing $\vec{b}$, the term $\vec{b} \cdot (\vec{a} 	imes \vec{b})$ is zero. Given $|\vec{b}| = 8$, the numerator simplifies to:

$$\vec{b} \cdot \vec{c} = -2(8)^2 = -128$$

### Final Calculation

Next, we determine the magnitude $|\vec{c}|$. Squaring the expression for $\vec{c}$ and noting that the cross term vanishes due to orthogonality:

$$|\vec{c}|^2 = (2\sqrt{2})^2 |\vec{a} 	imes \vec{b}|^2 + (-2)^2 |\vec{b}|^2$$

Substituting $|\vec{a} 	imes \vec{b}| = 4\sqrt{2}$ and $|\vec{b}| = 8$:

$$|\vec{c}|^2 = 8(32) + 4(64) = 256 + 256 = 512$$

Thus, $|\vec{c}| = \sqrt{512} = 16\sqrt{2}$. Substituting these values into our cosine formula:

$$\cos \alpha = \frac{-128}{8 	imes 16\sqrt{2}} = -\frac{1}{\sqrt{2}}$$

The angle $\alpha$ whose cosine is $-\frac{1}{\sqrt{2}}$ is **$\frac{3\pi}{4}$**.

Select Answer:

Visualized Solution

Visualizing the Parallelogram

Finding the Angle $θ$

Calculating Magnitude of $b$

Introducing Vector $c$

Calculating $b \cdot c$

Setting up Magnitude of $c$

Calculating Magnitude of $c$

Final Angle Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Trigonometric Harmony

The Construction of $c$

The Vanishing Act

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Parallelogram

Finding the Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Calculating Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Introducing Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b⋅c

Setting up Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Calculating Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c